Problem Description

Given an unsorted array of integers, find the length of the longest sequence of continuous (consecutive) numbers.

The numbers must be consecutive in value (like 1, 2, 3, 4) but do not need to be adjacent in the original array.

Example 1:

Input: arr = [100, 200, 4, 3, 1, 2]
Output: 4
Explanation: The longest consecutive sequence is [1, 2, 3, 4]

Example 2:

Input: arr = [1, 34, 35, 33, 23, 32, 11, 20]
Output: 4
Explanation: The longest consecutive sequence is [32, 33, 34, 35]

Approach

Sort the array first, then traverse it once while counting the current consecutive streak.

Key Insights:

1. After sorting, consecutive numbers appear next to each other
2. If arr[i] + 1 == arr[i + 1], increase the running streak
3. If duplicates appear, skip them without resetting the streak
4. For any gap, reset the streak to 1
5. Track the maximum streak seen so far

Solution

import java.util.Arrays;

class  Solution1{
    public static void main(String[] args){
        int[] arr = {100, 200, 4,3,1,2};
        int[] arr1 = {1, 34, 35,33,23,32,11,20};
        int n = returnLength(arr);
        System.out.println(n);

    }
    public static int returnLength(int[] arr){
        int n = arr.length;
        Arrays.sort(arr);
        int temp = 1;
        int count = 1;

        for(int i=0;i<n-1;i++){
            if(arr[i]+1==arr[i+1]){
                count++;
            }
            else if(arr[i]==arr[i+1]) {
                continue;
            }
            else{
                count = 1;
            }
            if(count>temp){
                temp = count;
            }
        }

        return temp;
    }

}

Time Complexity = O(n log n) Space Complexity = O(1)